## Derivation of the Expression for the Curl of a Vector Field in Cylindrical Coordinates

Let
$\mathbf{F} = F_r \mathbf{e_r} + F_{\theta} \mathbf{e_\theta} +F_z \mathbf{e_z}$
and let
$curl \mathbf{F} = \mathbf{\nabla} \times \mathbf{F} = F_A \mathbf{e_r} + A_{\theta} \mathbf{e_\theta} +A_z \mathbf{e_z}$

By definition,
$A_r = \frac{\oint_C \mathbf{F} \cdot d \mathbf{r}}{\int_S dS}$
where
$C$
is the boundary curve or the small surface element
$S$
on the curved surface of the cylinder shown.

\begin{aligned} \oint_C \mathbf{F} \cdot d \mathbf{r} &= \oint_{ABCD} \mathbf{F} d \mathbf{r} \\ &= \int_{AB} \mathbf{F} d \mathbf{r} + \int_{BC} \mathbf{F} d \mathbf{r} + \int_{CD} \mathbf{F} d \mathbf{r} + \int_{DA} \mathbf{F} d \mathbf{r} \\ &= (F_{\theta} r d \theta )_{AB} +(F_z dz)_{BC} - (F_{\theta} d \theta )_{CD} -(F_z dz)_{DA} \\ &= - \frac{\partial F_{\theta}}{\partial z}r dz d \theta + \frac{\partial F_z}{\partial \theta}d \theta dz \end{aligned}

The area of the surface element is
$r d \theta dz$
.
Hence
$A_r =\frac{ - \frac{\partial F_{\theta}}{\partial z}r dz d \theta + \frac{\partial F_z}{\partial \theta}d \theta dz}{r d \theta dz}= \frac{1}{r} \frac{\partial F_z}{\partial \theta} - \frac{\partial F_{\theta}}{\partial z}$

\begin{aligned} \oint_C \mathbf{F} \cdot d \mathbf{r} &= \oint_{EFGH} \mathbf{F} d \mathbf{r} \\ &= \int_{EF} \mathbf{F} d \mathbf{r} + \int_{FG} \mathbf{F} d \mathbf{r} + \int_{GH} \mathbf{F} d \mathbf{r} + \int_{HE} \mathbf{F} d \mathbf{r} \\ &= (F_z dz)_{EF} +(F_r dr)_{FG} - (F_z dz)_{GH} -(F_r dr)_{HE} \\ &= - \frac{\partial F_z}{\partial r}dr dz + \frac{\partial F_r}{\partial z}d r dz \end{aligned}

The area of the surface element is
$dr dz$
.
Hence
$A_\theta =\frac{- \frac{\partial F_z}{\partial r}dr dz + \frac{\partial F_r}{\partial z}d r dz}{dr dz}= \frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}$

\begin{aligned} \oint_C \mathbf{F} \cdot d \mathbf{r} &= \oint_{JKLM} \mathbf{F} d \mathbf{r} \\ &= \int_{JK} \mathbf{F} d \mathbf{r} + \int_{KL} \mathbf{F} d \mathbf{r} + \int_{LM} \mathbf{F} d \mathbf{r} + \int_{MJ} \mathbf{F} d \mathbf{r} \\ &= (F_r dr)_{JK} +(F_\theta d \theta)_{KL} - (F_r dr)_{LM} -(F_\theta d \theta)_{MJ} \\ &= - \frac{\partial F_r}{\partial \theta}dr d \theta + \frac{\partial (r F_\theta )}{\partial r}d r d \theta \end{aligned}

The area of the surface element is
$r d \theta dr$
.
Hence
$A_z =\frac{- \frac{\partial F_r}{\partial \theta}dr d \theta + \frac{\partial r F_\theta}{\partial r}d r d \theta}{r d \theta dr}= \frac{1}{r}(\frac{\partial (r F_\theta )}{\partial r} - \frac{\partial F_r}{\partial \theta})$

Hence
$\mathbf{\nabla} \times \mathbf{F} = (\frac{1}{r} \frac{\partial F_z}{\partial \theta} - \frac{\partial F_{\theta}}{\partial z}) \mathbf{e_r} + (\frac{\partial F_r}{\partial z} - \frac{\partial F_z}{\partial r}) \mathbf{e_\theta} + \frac{1}{r}(\frac{\partial (r F_\theta )}{\partial r} - \frac{\partial F_r}{\partial \theta}) \mathbf{e_z}$